The `x-t` graph of a particle undergoing simple harmonic motion is shown in figure. Acceleration of particle at `t = 4//3 s` is

Options

A. `(sqrt(3))/(32)pi^(2)cms^(-2)`
B. `-(pi^(2))/(32)cms^(-2)`
C. `(pi^(2))/(32)cms^(-2)`
D. `-(sqrt(3))/(32)pi^(2)s^(-2)`

Answer

D

Explanation

From graph,
`A=1cm,T=8s`
`x=Asinomegat=A"sin"(2pi)/(T)t`
At `t=(4)/(3)s`,
`x=1"sin"(2pi)/(8)xx(4)/(3)="sin"(pi)/(3)=(sqrt(3))/(2)cm`
In SHM,
Acceleration`=-omega^(2)x`
`=-(4pi^(2))/(T^(2))xx(sqrt(3))/(2)" cm "s^(-2)" "(becauseomega=(2pi)/(T))`
`=-(4pi^(2))/((8)^(2))xx(sqrt(3))/(2)=-(sqrt(3)pi^(2))/(32)cm" "s^(-2)`